Answer:
import java.util.Scanner;
public class InputExample {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int birthMonth;
int birthYear;
birthMonth = scnr.nextInt();
birthYear = scnr.nextInt();
System.out.println(birthMonth+"/"+birthYear);
}
}
Web developers design and create websites. They are responsible for the look of the site. They are also responsible for the site's technical aspects, such as its performance and capacity, which are measures of a website's speed and how much traffic the site can handle. In addition, web developers may create content for the site
Answer:
Explanation:
Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F
To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.
where i=1 to N-1 and N=6
A area of the polygon can be found by the following formula
where i=1 to N-1
![A=\frac{1}{2}[ (x_{1} y_{2} -x_{2} y_{1})+ (x_{2} y_{3} -x_{3} y_{2})+(x_{3} y_{4} -x_{4} y_{3})+(x_{4} y_{5} -x_{5} y_{4})+(x_{5} y_{6} -x_{6} y_{5})]](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%5B%20%28x_%7B1%7D%20%20y_%7B2%7D%20-x_%7B2%7D%20%20y_%7B1%7D%29%2B%20%28x_%7B2%7D%20%20y_%7B3%7D%20-x_%7B3%7D%20%20y_%7B2%7D%29%2B%28x_%7B3%7D%20%20y_%7B4%7D%20-x_%7B4%7D%20%20y_%7B3%7D%29%2B%28x_%7B4%7D%20%20y_%7B5%7D%20-x_%7B5%7D%20%20y_%7B4%7D%29%2B%28x_%7B5%7D%20%20y_%7B6%7D%20-x_%7B6%7D%20%20y_%7B5%7D%29%5D)
A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)
A=225.5 miles²
Now putting the value of area in Cx and Cy
![C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(x_{2}+x_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(x_{3}+x_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(x_{4}+x_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(x_{5}+x_{6})(x_{5} y_{6} -x_{6} y_{5})]]](https://tex.z-dn.net/?f=C_%7Bx%7D%20%3D%5Cfrac%7B1%7D%7B6A%7D%5B%20%5B%28x_%7B1%7D%2Bx_%7B2%7D%29%28x_%7B1%7D%20%20y_%7B2%7D%20-x_%7B2%7D%20%20y_%7B1%7D%29%5D%2B%20%5B%28x_%7B2%7D%2Bx_%7B3%7D%29%28x_%7B2%7D%20%20y_%7B3%7D%20-x_%7B3%7D%20%20y_%7B2%7D%29%5D%2B%5B%28x_%7B3%7D%2Bx_%7B4%7D%29%28x_%7B3%7D%20%20y_%7B4%7D%20-x_%7B4%7D%20%20y_%7B3%7D%29%5D%2B%5B%28x_%7B4%7D%2Bx_%7B5%7D%29%28x_%7B4%7D%20%20y_%7B5%7D%20-x_%7B5%7D%20%20y_%7B4%7D%29%5D%2B%5B%28x_%7B5%7D%2Bx_%7B6%7D%29%28x_%7B5%7D%20%20y_%7B6%7D%20-x_%7B6%7D%20%20y_%7B5%7D%29%5D%5D)
putting the values of x's and y's you will get

For Cy
![C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(y_{2}+y_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(y_{3}+y_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(y_{4}+y_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(y_{5}+y_{6})(x_{5} y_{6} -x_{6} y_{5})]]](https://tex.z-dn.net/?f=C_%7By%7D%20%3D%5Cfrac%7B1%7D%7B6A%7D%5B%20%5B%28y_%7B1%7D%2By_%7B2%7D%29%28x_%7B1%7D%20%20y_%7B2%7D%20-x_%7B2%7D%20%20y_%7B1%7D%29%5D%2B%20%5B%28y_%7B2%7D%2By_%7B3%7D%29%28x_%7B2%7D%20%20y_%7B3%7D%20-x_%7B3%7D%20%20y_%7B2%7D%29%5D%2B%5B%28y_%7B3%7D%2By_%7B4%7D%29%28x_%7B3%7D%20%20y_%7B4%7D%20-x_%7B4%7D%20%20y_%7B3%7D%29%5D%2B%5B%28y_%7B4%7D%2By_%7B5%7D%29%28x_%7B4%7D%20%20y_%7B5%7D%20-x_%7B5%7D%20%20y_%7B4%7D%29%5D%2B%5B%28y_%7B5%7D%2By_%7B6%7D%29%28x_%7B5%7D%20%20y_%7B6%7D%20-x_%7B6%7D%20%20y_%7B5%7D%29%5D%5D)
putting the values of x's and y's you will get

So coordinates for the fire station should be (15.36,22.55)
Answer:

Explanation:
The temperature can be defined as the measurement of the intensity of the heat present in the object. Fahrenheit, kelvin and centigrade are the common scale used for measuring Temperature.
Given:
T1=170C
To convert to Kelvin
= 17+273 =290K
T1 = 290K
Pressure (P)= 95KPa
Specific heat ratio = CP/CV= K
WhereK=1.005/0.718
K = 1.4
The final temperature can be calculated using the formula below.
T2 = CP/CV × T1
=. K × T1
T2 = 1.4 × 290

Answer:
I can help but I need to know what it looking for